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Building a micro (ceramic) heating element for vaporizing e-liquid? Answered

Hi there, 

I'm designing a vaping device which is Li-Ion battery powered and uses standard e-liquids (propylene glycol / vegetable glycol). The liquid would evaporate around 200 celsius and the heating element should attain this temperature within 1 second. Current methods use a coil with a liquid absorbing wire, which is not durable nor easy to reuse. 

I've been looking into ceramic heating elements but it's a forest of options out there, most of them becoming waaaaay to hot and/or needing a much higher voltage than a standaard Li-ion battery would provide at around 1100 Mah. So i've been thinking of building one, or having a prototype built for me. Problem is, i'm not 100% sure that a ceramic method would be the best way (so many options). 

Around the ceramic element (or on top if it is in disc form), i'd place a replaceable sock that can absorb the fluid and withstand excess heat (better than the standard kanthal / ekowool / etc.). This way the end user would only need to replace a sock after some time (for instance when changing fluids). It would also not burn and contaminate the liquid like current system do. 

Anybody have an idea? 




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Jack A Lopez
Jack A Lopez

6 years ago

This is a question that has been asked before, sort of. I mean, previously in this forum I have seen questions like, "So I want to build a heating element, and I want it to reach a temperature of so many degrees Celcius..."

And it is a problem people have though about before, for example, here:


I know you want some specific temperature, like 200 C. However from the perspective of electrical design, it is much easier to design a heating element capable of dissipating some specific amount of power, in watts, and the trick to that is to just assume my heating element is a resistor. If I have some amount of power P, in mind, I can use the formulas for power and Ohm's law, P = I*V = V^2/R = I^2*R, to find a resistance R that is a good match to my power supply.

E.g. if I've got a 5 volt DC supply, and I want to dissipate 10 watts of heat, then a 2.5 ohm resistor would do exactly that. Since P= V^2/R = (5V*5V)/(2.5 ohm) = 10 W

So maybe now you are wondering how much power you need? I mean, for to heat the inside of a small oven-like enclosure to the temperature you want.

Unfortunately, there is not a simple relationship between input power (in watts) and the hottest temperature (in degrees C) in the center of the oven. Mostly it depends on how fast heat is flowing out of the hot center.

As an extreme example, consider the energy that flows into, and out of, the filament in an incandescent flashlight bulb. The actual power flow, electricty in, or heat and light out, is going to be approximately 1 watt. Yet the temperature of the filament is yellow hot, something like 3000 C. The filament is tiny, and it is surrounded by a good insulator, gas that is almost vacuum. For this reason heat is basically "trapped" in the filament, and the temperature there can get really hot.

As a contrasting example, consider a pot of water, on an electric hot plate. The power flow into something like that could be hundreds of watts, yet it takes several minutes for the water to get hot enough to boil (circa 100 C). Something like this does not get hot quickly, despite huge amounts of energy being poured into it. The reason why is the heat has lots of places to go. Also once the water starts boiling the heat has places to go and things to do, things like turning the water into steam.

Similarly, for your thing, heat is flowing out in two ways. One way is just the usual heat leaking through the walls of the oven. The second way heat is being lost is in the heat needed to vaporize your liquid, the heat needed for the actual function of your thing.

I think the way forward is to just sort of take wild guess


at how much power (in watts) you need for to heat your little oven thing, and then pick a suitable sized resistor, or hand-made resistor made from nichrome wire, then build it, and actually measure its temperature using a thermocouple thermometer, and also measure the power flowing in using your electrical measuring tools, e.g. a voltmeter and an ammeter.

If you have some way of adjusting the input power, e.g. by using pulses of variable width (also called PWM) instead of just straight DC, that might be helpful.

Also if you have the skill for it, you could build an oven that is thermostatically controlled. That is to say you have some temperature sensor, like a thermocouple, producing an electric signal, and that signal can be used as feedback to determine how much PWM, how much input power is needed to maintain that temperature.


Answer 6 years ago

Hi Jack,

thanx for the input, very clarifying as i'm not an engineer. But consider that I can get a heating element of about 5x5mm (that should be enough), wouldn't you say that, in general, this wouldn't need a lot of power? It doesn't have to boil water, heating a segment of a fluid absorbing sock of that size is enough (and the width of the sock would only be 1-2 mm).

Jack A Lopez
Jack A Lopez

Answer 6 years ago

I just noticed one of those images was from a YouTube video, some somebody reviewing the Pilot(r) brand Slider(r) cigarette lighter. From looking at the packaging of the lighter reviewed in this video, I am pretty sure it is the same thing I saw on the street today. Anyway, watch this vid, and you'll know as much as I do about this electric lighter gizmo, or actually you'll probably know more about it, since I didn't have the patience to watch this vid all the way through.


Jack A Lopez
Jack A Lopez

Answer 6 years ago

Hello Bert,

You are correct in thinking that heating a small volume of something will take much less power than heating a large volume.

I don't know what part of the world you live in. I live in the former United States. The reason this is relevant to the discussion is because coincidentally, today I was waiting in line the local WalMart(tm) (a big-box retailer in the FUS) when I noticed that, among all the other junk in this check out aisle, that they were selling some sort of electric, USB-rechargeable, cigarette lighter, at a price of about 9 USD.

Reading the blurb on the back of the card revealed it was powered by some kind of lithium, or lithium polymer, battery.

It was also branded as a "Slider", "USB", something, something, and it turns out that words like this are enough for Google(tm) images to show me pictures of similar things, here:

The reason this is relevant, is that if you can find one of these artifacts, for sale on the street, wherever you live, such an artifact already contains many of the parts you are interested in, specifically: a rechargeable lithium battery, plus a small heating element which is well-matched, ie. made to work with, the voltage-current characteristic of this battery.

Coincidentally the area of this tiny heating element, the one in the "Slider-USB" cigarette lighter gizmo, is about the cross-section of a cigarette, which is roughly a circle about 6 mm in diameter, or pi*6mm^2 = 30 mm^2, which is roughly the same size as you say you want, about 5mmx5mm.

Guessing the actual amount of raw power dissipated by that gizmo is a few watts, about 3 volts multiplied by about 0.5 - 2.0 amperes, and that's a wild-ass guess.

To summarize: I think what you want to do is possible, and I might have even seen a comparable implementation of it while standing in line at the supermarket today.

BTW, sorry it took like four days to get an answer to your question, but this forum is total voluntary anarchy, so you kind of get what you get.